Discussion
The inverse Laplace Transform of \(\dfrac{s^{2}}{(s-3)^{3}}\) can be written as \(\dfrac{e^{3t}}{2}[At^{2} + Bt + C]\). The values of A, B and C, respectively are
Answer: D
The values are A = 9, B = 12 and C = 2.
Write \(s^{2} = (s-3)^{2} + 6(s-3) + 9\) and split the fraction, then use \(\mathcal{L}^{-1}\{1/(s-a)^{k}\} = e^{at}t^{k-1}/(k-1)!\):
\[ \begin{aligned} \frac{s^{2}}{(s-3)^{3}} &= \frac{1}{s-3} + \frac{6}{(s-3)^{2}} + \frac{9}{(s-3)^{3}} \\ f(t) &= e^{3t}\left(1 + 6t + \tfrac{9}{2}t^{2}\right) \\ &= \frac{e^{3t}}{2}\left[9t^{2} + 12t + 2\right] \end{aligned} \]
Check at t = 0: the initial value theorem gives \(\lim s \cdot s^{2}/(s-3)^{3} = 1\), and \(C/2 = 1\) needs C = 2, which only the keyed option has.
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